Calculate the potential for the reaction below at 25ºC under the
following conditions:
[Fe+3 ]=0.0200M
[Sn+2 ]=0.0200M
[Fe+2 ]=1.50M
[Sn+4 ]=1.50M
2 Fe+3(aq) + Sn+2 (aq) → 2 Fe+2
(aq) + Sn+4 (aq)
First calculate standard cell potential using standard reduction potential.
Calculate Qr
Put all in nearnst equation.. you will get the answer.

Calculate the potential for the reaction below at 25ºC under the following conditions: [Fe+3 ]=0.0200M [Sn+2...
Calculate the potential for the reaction below at 25ºC under the following conditions: [Fe+3 ]=0.0200M [Sn+2 ]=0.0200M [Fe+2 ]=1.50M [Sn+4 ]=1.50M 2 Fe+3(aq) + Sn+2 (aq) → 2 Fe+2 (aq) + Sn+4 (aq) E = Answer V
a)Calculate the potential for the reaction below at 25ºC under the following conditions: [MnO?¯ ]=0.0200M [Ni+2 ]=2.50M pH = 2.000 2 MnO?¯ (aq) + 8 H+ (aq) + 3 Ni (s) ? 2 MnO? (s) + 4 H?O (l) + 3 Ni+2 (aq)
u Calculate the standard cell potential for each reaction below, andnote whether the reaction is spontaneous under standard state conditions. 1. Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s) 2. Na(s)+LiNO3(aq)⟶NaNO3(aq)+Li(s) ● 3. Mg(?)+Ni2+(??)⟶Mg2+(??)+Ni(?)
Calculate the standard potential for the reaction shown below at 25º? Sn+4 (aq) + Cl – (aq) + 2 OH – (aq) ⟶ Sn+2 (aq) + ClO – (aq) + H2O (l) Eº = Answer V
the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) fone 2 pts) a. Use the standard half-cell potentials listed below to calculate the standard cell potential (Eºcell) for 3 Sn(s) + 2 Fe* (aq) - 3 Sn2+ (aq) + 2 Fe(s) Sn 2(aq) + 2 e -Sn (s) E = -0.14 V Fe3+ (aq) + 3 e Fe(s) E° = -0.036 V A) -0.176 V B)-0.104 V C) +0.104 V D) +0.176 V b. Write...
Calculate the standard cell potential for the following reaction. Fe (s) +Ni^+2 (aq) rightarrow Fe^+ 2 (aq) + Ni (s) 3 Cu + 2 NO_3^- + 8 H^+ rightarrow 3 Cu^+ 2+ 2 NO +4 H_2O Cr_2O_7^-2 + 6 Fe^+2+14 H6+ rightarrow 2 Cr^+ 3 + 6 Fe^+ 3 +7 H_2O
Under the conditions shown below for the following reaction. Fe2O3(s) + 3 CO(g) - 2 Fe(s) + 3 CO2(g) AG° = -28.0 kJ P(CO) - 3.1 atm, P(CO2) - 1.7 atm Calculate Arxn at 298 K, and indicate if the reaction is more or less spontaneous under these conditions than under standard conditions? -4.5 kJ, more spontaneous than under standard conditions None of these -4.5 kJ, less spontaneous than under standard conditions -23.5 kJ, more spontaneous than under standard conditions...
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
1) Balance the following reaction under acidic conditions and calculate the cell potential in (V) at 298 K generated by the cell. Report your answer to the hundredths place. Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s) [Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M, [Cr3+] = 1.0 × 10-5 M 2) What is the value of n for the following reaction? Enter the whole number. 3Ni+(aq) + Cr(OH)3(s) + 5OH-(aq) → 3Ni(s) + CrO42-(aq) + 4H2O(l)
A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)---- Sn(s)+Mn2+(aq) Calculate the cell potential of 25 degrees Celsius under each of the following conditions. Part A: Sn2+= 1.15*10^-2 M; and Mn2+= 2.37 M Part B: Sn2+= 2.37 M; and Mn2+= 1.15*10^-2