1. Cu|Cu^ 2+ (1M)||Ag^+(1M)|Ag E^0_ Cu^ 2+ Cu =0.34V E^0_ Ag^+ Ag… - Vidyadip

Question

$\text{Cu}|\text{Cu}^{2+}(1\text{M})||\text{Ag}^+(1\text{M})|\text{Ag}$

$\text{E}^0_\frac{\text{Cu}^{2+}}{\text{Cu}}=0.34\text{V}$

$\text{E}^0_\frac{\text{Ag}^+}{\text{Ag}}=+0.80\text{V}$

Write reaction at anode and cathode.
Write net cell reaction.
Calculate e.m.f. of the cell.

What is meant by E.M.F of cell? How is it measured?

✓

Answer

$\text{Cu}\text{(S)}\xrightarrow{ \ \ \ \ }\text{Cu}^{2+}\text{(aq)}+2\text{e}^-$ At anode

$2\text{Ag}^+\text{(aq)}+2\text{e}^-\xrightarrow{ \ \ \ \ \ \ }2\text{Ag}\text{(s)}$ At cathode

$\text{Cu}\text{(s)}+2\text{Ag}^+\text{(aq)}\xrightarrow{ \ \ \ \ \ }\text{Cu}^{2+}\text{(aq)}+2\text{Ag(s)}$

$\text{E}^0_\text{cell}=\text{E}^0_\frac{\text{Ag}+}{\text{Ag}}-\text{E}^0_\frac{\text{Cu}^{2+}}{\text{Cu}}$

$=+0.80\text{V}-0.34\text{V}$

$=0.46\text{V}$

It is equal to maximum potential difference when no current is drawn from the cell.

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