Determine Ecell (in V) at 298.15 K for a cell composed of a platinum electrode in a mixture of 0.629 M Fe2+and 0.163 M Fe3+ coupled to a platinum electrode in a solution where the [H+] is 1.546 M and the partial pressure of H2 is 0.709 atm.
to solve this problem we need the standats reduction potential
of Fe+3/Fe+2 which is .77 volt while standard reduction potential
of hydrogen is 0.0 volt and then applying nerst equation.
Determine Ecell (in V) at 298.15 K for a cell composed of a platinum electrode in...
a) Determine Ecell (in V) for a cell composed of a Ag electrode in a solution of Ag+ coupled to a standard hydrogen electrode (SHE) under standard conditions. Report your answer to three decimal places in standard notation (i.e. 1.234 V) Ag+(aq) + e− ⇌ Ag(s) E° = 0.800 V b) Determine Ecell (in V) at 298.15 K for a cell composed of a Ag electrode in a solution of 0.528 M Ag+ coupled to a platinum electrode in a...
0.741 V You are correct. Your receipt no. is 156-8240 Previous Tries 2. Determine Ecell (in V) at 298.15 K for a cell composed of a Ti electrode in a solution of 0.372 M + coupled to a platinum electrode in a solution where the [H] is 0.984 M and the partial pressure of Hy is 1.500 atm. Report your answer to three decimal places in standard notation (I.e. 1.234 V).
1. Use the following standard cell potentials to answer the questions below. Yb3+(aq) + 3e− ⇌ Yb(s) E° = -2.190 V Au3+(aq) + 2e− ⇌ Au+(aq) E° = 1.401 V (a) What is ΔG° (in kJ/mol) for this reaction? Report answer to three significant figures in scientific notation (i.e., 1.23e2 kJ/mol) (b) ) This reaction is (spontaneous or not spontaneous). 2. Determine Ecell (in V) at 298.15 K for a cell composed of a platinum electrode in a mixture of...
this is how to question is give.
A cell composed of a platinum indicator electrode and a silver-silver chloride reference electrode in a solution containing both Fe2* and Fe3+ has a cell potential of 0.699 V. If the silver-silver chloride electrode is replaced with a saturated calomel electrode (SCE), what is the new cell potential? Number 0.675
V3+(aq) + e− ⇌ V2+(aq) E° = -0.255 V 2H+(aq) + 2e− ⇌ H2(g) E° = 0.000 V 3. The electrochemical cell is comprised of a Pt electrode in a 4.16 × 10-4 M solution of V3+ and 7.29 × 10-2 M solution of V2+ coupled to a Pt electrode where the [H+] is 8.67 × 10-5 M and the partial pressure of H2(g) is 0.690 atm. The temperature of this cell is held constant at 298.15 K (a) Under...
2. The electrochemical cell is comprised of a Mo electrode in a 3.87 × 10-1 M solution of Mo3+ (aq) coupled to a Pt electrode in a solution containing H+ (aq) where the pH of the solution is 0.16 and the partial pressure of H2(g) is 1.018 atm. The temperature of the cell is held constant at 25°C. (a)What is Ecell (in V) for the electrochemical cell? Report your answer to three decimal places in standard notation (i.e., 0.123 V)...
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2 half- cell and an H2/H half-cell under the following conditions: [Zn2] = 0.042 M [H]- 19 M partial pressure of H2 =0.37 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2 (aq) + 2e +2H (aq) + 2e1 Eo-0.76 V E 0.00 V Zn(s) H2(g)
In a test of a new reference electrode, a chemist constructs a
voltaic cell consisting of a Zn/Zn2+ half-cell and an
H2/H+ half-cell under the following
conditions: [Zn2+ ] = 0.021 M [H+ ]= 1.3 M
partial pressure of H2 = 0.32 atm. Calculate
Ecell at 298 K (enter to 3 decimal places).
Zn2+ (aq) + 2e −
⟶ Zn(s) E° = − 0.76 V
2H+ (aq) + 2e −
⟶ H2(g) E° = 0.00 V
We were unable...
Pb2+(aq) + 2e− ⇌ Pb(s) E° = -0.126 V 2H+(aq) + 2e− ⇌ H2(g) E° = 0.000 V E°cell (in V)= 0.126 V 2. The electrochemical cell is comprised of a Pb electrode in a 1.67 × 100 M solution of Pb2+ (aq) coupled to a Pt electrode in a solution containing H+ (aq) where the pH of the solution is 0.37 and the partial pressure of H2(g) is 0.571 atm. The temperature of the cell is held constant at...
Consider the cell Pt(s)|H2(g,1atm)|H+(aq,a=1)|Fe3+(aq),Fe2+(aq)|Pt(s) given that Fe3++e−⇌Fe2+ and E∘=0.771V at 298.15 K. If the cell potential is 0.683 V, what is the ratio of Fe2+(aq) to Fe3+(aq)? What is the ratio of these concentrations if the cell potential is 0.807 V?