First note that the Ni electrode is not in 1 M solution; therefore, the E value is not Eo. That must be corrected.
E = Eored - 0.RT/nF*[log (red form)/(oxidized form)]
Plug in the values. -0.230 - (0.0592/2)log (Ni)/(Ni^2+). Plug in 1 for Ni and 0.0696 for (Ni^2+) and solve for Ered. Reverse that for Eox.
Ag^+ + e ==> Ag(s) Eored = +0.230
Ni ==> Ni^2+ + 2e Eoox = value from above.
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2Ag + Ni(s) => 2Ag(s) + Ni^2+ and Ecell = Ered + Eox
Note also: The values I found for E for Ag/AgCl on the web were not consistent. Use the values in your text or notes for this problem and not the numbers I've used.
A galvanic cell is constructed with a silver-silver chloride electrode, and a nickel strip immersed in a beaker containing 6.96 x 10-2 M solution of NiCl2. Determine the balanced cell reaction and calculate the potential of the cell. Enter in volts. (assume a temperature of 25°C)
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