Question

Calculate the EMF between copper and silver Ag+e-E=0.89v
Cu=E=0.34v​

Answer 1

Depending on the
`E^\circ` value of
`\rm Ag^(+) + e^(-) \to Ag\; (s)`, the cell potential would be:

• 
  `0.55\; \rm V`, using data from this particular question; or
• approximately
  `0.46\; \rm V`, using data from the CRC handbooks.

Step-by-step explanation:

In this galvanic cell, the following two reactions are going on:

• The conversion between
  `\rm Ag\; (s)` and
  `\rm Ag^(+)` ions,
  `\rm Ag^(+) + e^(-) \rightleftharpoons Ag\; (s)`, and
• The conversion between
  `\rm Cu\; (s)` and
  `\rm Cu^(2+)` ions,
  `\rm Cu^(2+)\; (aq) + 2\, e^(-) \rightleftharpoons \rm Cu\; (s)`.

Note that the standard reduction potential of
`\rm Ag^(+)` ions to
`\rm Ag\; (s)` is higher than that of
`\rm Cu^(2+)` ions to
`\rm Cu\; (s)`. Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if
`\rm Ag^(+)` ions are reduced while
`\rm Cu\; (s)` is oxidized.

Therefore:

• The reduction reaction at the cathode will be:
  `\rm Ag^(+) + e^(-) \to Ag\; (s)`. The standard cell potential of this reaction (according to this question) is
  `E(\text{cathode}) = 0.89\; \rm V`. According to the 2012 CRC handbook, that value will be approximately
  `0.79\; \rm V`.

• The oxidation at the anode will be:
  `\rm Cu\; (s) \to \rm Cu^(2+) + 2\, e^(-)`. According to this question, this reaction in the opposite direction (
  `\rm Cu^(2+)\; (aq) + 2\, e^(-) \rightleftharpoons \rm Cu\; (s)`) has an electrode potential of
  `0.34\; \rm V`. When that reaction is inverted, the electrode potential will also be inverted. Therefore,
  `E(\text{anode}) = -0.34\; \rm V`.

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

`\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}`.

Using data from the 1985 and 2012 CRC Handbook:

`\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}`.