What is the expected standard cell potential for the following unbalanced reaction under basic conditions? MnO4-(aq)...
What is the expected standard cell potential for the following unbalanced reaction under basic conditions? MnO4-(aq) + Fe+2(aq) ⟶ Fe+3 (aq) + MnO2(s) Given: Fe3+ + e− ⟶ Fe2+ E1/2= 0.771 MnO4−+ 2 H2O + 3e− ⟶ MnO2 + 4OH− E1/2 = 0.558 A. -0.213 V B. +0.213 V C. -1.329 V D. +1.329 V E. None of the above
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What is the expected standard cell potential for the following
unbalanced reaction under basic conditions? MnO4-(aq)...
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction...
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe(s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn(aq) +2Cr(s) →2Cr" (aq) +35n²+ (aq)
4. Calculate the cell potential for the half cells Fe3+/Fe2+ and MnO4- /Mn2+, where the Mn...
4. Calculate the cell potential for the half cells Fe3+/Fe2+ and MnO4- /Mn2+, where the Mn process occurs at the cathode, under the following conditions and predict whether the reaction is spontaneous: [Fe3+] = 1.0 M, [Fe2+] = 0.1 M, [MnO4 - ] = 0.01 M [Mn2+] = 1 x 10-4 M [H+ ] = 1 x 10-3 M Fe3+ + e− → Fe2+ +0.700 V MnO4 − (aq) + 8 H+ (aq)+ 5e− → Mn2+ (aq) + 4 H2O(l)...
A voltaic cell utilizes the following reaction: 2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq). emf of this cell under standard conditions E∘...
A voltaic cell utilizes the following reaction: 2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq). emf of this cell under standard conditions E∘ = 0.771 V What is the emf for this cell when [Fe3+]= 3.70 M , PH2= 0.95 atm , [Fe2+]= 1.0×10−3 M , and the pH in both compartments is 3.95? Express your answer using two significant figures.
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction +...
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
What is the standard cell potential for a voltaic cell with the following line notation? Al(s)[A1+...
What is the standard cell potential for a voltaic cell with the following line notation? Al(s)[A1+ (aq)||Fe3+ (aq) Fe(s) Give your answer to 3 significant figures. Half Reaction Eredº (V) Al3+ + 3e- → AI -1.66 Fe3+ + 3e- → Fe +0.771 Numeric Answer
When the following reaction is balanced under basic conditions, what is the ratio of the coefficients...
When the following reaction is balanced under basic conditions, what is the ratio of the coefficients of Mn(OH)2(s) to MnO4--(aq)? Mn(OH)2(s) + MnO4 (aq) + MnO42-(aq) (A) 3:1 (B) 1:3 (C) 1:4 (D) 1:5 What is the standard reduction potential for the reduction of permanganate ion to managanese dioxide in acidic solution? Half-Reaction E. V MnO4 (aq) + 8 H(aq) + 5 € → Mn²+ (aq) + +1.51 4 H2O(1) MnO2(s) + 4 H (aq) + 2 e → Mn2+(aq)...
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5)...
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5) Sn+ (aq) + 2e - Sn2(aq) E (V) -0.74 -0.440 +0.771 +0.154 1. Calculate the standard cell potential for the voltaic cell based on the reaction below, given the table above: 35nt(s) + 2Cr (s) - 2 C (s) + 3 Sn () ANSWER: 2. Calculate the standard cell potential for the voltaic cell based on the reaction below, Riven the table above 3Feb...
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr (s) -0.74 Fe2+ (aq) +...
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr (s) -0.74 Fe2+ (aq) + 2e- Fe () -0.440 Fe3+ (aq) + e- → Fe2+ (s) +0.771 Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 The standard cell potential (Eºcell) for the voltaic cell based on the reaction below is V. Cr (s) + 3Fe3+ (aq) + 3Fe2+ (aq) + Cr3+ (aq) A) -1.45 B) +2.99 C) +1.51 D) +3.05 E) +1.57
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr(s) -0.74 Fe2+ (aq) + 2e- Fe (5) -0.440 Fe3+ (aq) + e - Fe2+ (s...
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr(s) -0.74 Fe2+ (aq) + 2e- Fe (5) -0.440 Fe3+ (aq) + e - Fe2+ (s) +0.771 Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 The standard cell potential (Eºcell) for the voltaic cell based on the reaction below is V. 35n4+ (aq) + 2Cr (s) → 2Cr3+ (aq) + 3Sn2+ (aq) A) +1.94 B) +0.89 C) +2.53 D) -0.59 E) -1.02
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq...
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn** (aq) +2Cr(s) → 2Cr** (aq) +3Sn²+ (aq) 3. What is the cell potential, Ecell, for the reaction above if [Sn] = 1.00 M, [Cr3+1 = 0.0200 M and [Sn2+] = 0.0100 M?
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe(s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn(aq) +2Cr(s) →2Cr" (aq) +35n²+ (aq)
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
What is the standard cell potential for a voltaic cell with the following line notation? Al(s)[A1+ (aq)||Fe3+ (aq) Fe(s) Give your answer to 3 significant figures. Half Reaction Eredº (V) Al3+ + 3e- → AI -1.66 Fe3+ + 3e- → Fe +0.771 Numeric Answer
When the following reaction is balanced under basic conditions, what is the ratio of the coefficients of Mn(OH)2(s) to MnO4--(aq)? Mn(OH)2(s) + MnO4 (aq) + MnO42-(aq) (A) 3:1 (B) 1:3 (C) 1:4 (D) 1:5 What is the standard reduction potential for the reduction of permanganate ion to managanese dioxide in acidic solution? Half-Reaction E. V MnO4 (aq) + 8 H(aq) + 5 € → Mn²+ (aq) + +1.51 4 H2O(1) MnO2(s) + 4 H (aq) + 2 e → Mn2+(aq)...
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5) Sn+ (aq) + 2e - Sn2(aq) E (V) -0.74 -0.440 +0.771 +0.154 1. Calculate the standard cell potential for the voltaic cell based on the reaction below, given the table above: 35nt(s) + 2Cr (s) - 2 C (s) + 3 Sn () ANSWER: 2. Calculate the standard cell potential for the voltaic cell based on the reaction below, Riven the table above 3Feb...
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr (s) -0.74 Fe2+ (aq) + 2e- Fe () -0.440 Fe3+ (aq) + e- → Fe2+ (s) +0.771 Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 The standard cell potential (Eºcell) for the voltaic cell based on the reaction below is V. Cr (s) + 3Fe3+ (aq) + 3Fe2+ (aq) + Cr3+ (aq) A) -1.45 B) +2.99 C) +1.51 D) +3.05 E) +1.57
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr(s) -0.74 Fe2+ (aq) + 2e- Fe (5) -0.440 Fe3+ (aq) + e - Fe2+ (s) +0.771 Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 The standard cell potential (Eºcell) for the voltaic cell based on the reaction below is V. 35n4+ (aq) + 2Cr (s) → 2Cr3+ (aq) + 3Sn2+ (aq) A) +1.94 B) +0.89 C) +2.53 D) -0.59 E) -1.02
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn** (aq) +2Cr(s) → 2Cr** (aq) +3Sn²+ (aq) 3. What is the cell potential, Ecell, for the reaction above if [Sn] = 1.00 M, [Cr3+1 = 0.0200 M and [Sn2+] = 0.0100 M?
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