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Electrochemistry — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
A copper-silver cell is set up. The copper ion concentration in it is $0.10 M$. The concentration of silver ion is not known. The cell potential measured $0.422 V$. Determine the concentration of silver ion in the cell.
$\text{Given: E}^{o}_{\text{Ag}^{+}/\text{Ag}}=+0.80\text{V},\text{E}^{o}_{\text{Cu}^{2+}/\text{Cu}}=+0.34\text{V}.$
✓

Answer

The cell reaction: $Cu(s) + 2 Ag^+(aq) → Cu^{2+}(aq) + 2 Ag(s)$
$\text{E}^{\circ}_{\text{cell}}=0.46\text{ V}$
Nernst equation
$\text{E}_{\text{cell}}=\text{E}^{\circ}_{\text{cell}}-\frac{0.059}{2}\log\frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^{2}}$
$\text{E}_{\text{cell}}=(0.80-0.34)-\frac{0.059}{2}\log\frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^{2}}$
$0.422\text{V}=0.46\text{V}-\frac{0.059}{2}\log\frac{0.10}{[\text{Ag}^{+}]^{2}}$
$\log\frac{0.10}{[\text{Ag}^{+}]^{2}}=1.2881$
$[Ag^+]^2 = 0.0051$
$[Ag^+] = 7.1 x 10^{–2} M.$

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