Solved: A voltaic cell is constructed from an Ni2+(aq) –
Chapter , Problem 101AE(choose chapter or problem)
Problem 101AE
A voltaic cell is constructed from an Ni2+(aq) – Ni(s) half-cell and an Ag+(aq) - Ag(s) half-cell. The initial concentration of Ni2+(aq) in the Ni2+ - Ni half-cell is [Ni2+] = 0.0100 M. The initial cell voltage is +1.12 V.
(a) By using data in Table, calculate the standard emf of this voltaic cell.
Table Standard Reduction Potentials in Water at 25°C
E°red(V) |
Reduction Half-Reaction |
+2.87 |
F2(g) + 2 e-→ 2 F-(aq) |
+1.51 |
MnO4-(aq) + 8 H+(aq) + 5 e-→ Mn2+(aq) + 4 H2O(l) |
+1.36 |
Cl2(g) + 2 e-→ 2 Cl-(aq) |
+1.33 |
Cr2O72-(aq) + 14 H+(aq) + 6 e-→ 2 Cr3+(aq) + 7 H2O(l) |
+1.23 |
O2(g) + 4 H+(aq) + 4 e-→ 2 H2O(l) |
+1.06 |
Br2(l) + 2 e-→ 2 Br-(aq) |
+0.96 |
NO3-(aq) + 4 H+(aq) + 3 e-→ NO(g) + 2 H2O(l) |
+0.80 |
Ag+(aq) + e-→ Ag(s) |
+0.77 |
Fe3+(aq) + e-→ Fe2+(aq) |
+0.68 |
O2(g) + 2 H+(aq) + 2 e-→ H2O2(aq) |
+0.59 |
MnO4-(aq) + 2 H2O(l) + 3 e-→ MnO2(s) + 4 OH-(aq) |
+0.54 |
l2(s) + 2 e-→ 2 l-(aq) |
+0.40 |
O2(g) + 2 H2O(l) + 4 e-→ 4 OH-(aq) |
+0.34 |
Cu2+(aq) + 2 e-→ Cu(s) |
0 [defined] |
2 H+(aq) + 2 e-→ H2(g) |
-0.28 |
Ni2+(aq) + 2 e-→ Ni(s) |
-0.44 |
Fe2+ (aq) + 2 e-→ Fe(s) |
-0.76 |
Zn2+(aq) + 2 e-→ Zn(s) |
-0.83 |
2 H2O(l) + 2 e-→ H2(g) + 2 OH-(aq) |
-1.66 |
Al3+(aq) + 3 e-→ Al(s) |
-2.71 |
Na+(aq) + e-→ Na(s) |
-3.05 |
Li+(aq) + e-→ Li(s) |
(b) Will the concentration of Ni2+(aq) increase or decrease as the cell operates? (c) What is the initial concentration of Ag+(aq) in the Ag+-Ag half-cell?
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