JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The cell potential for the given cell at \(298 \,K\) \(Pt\) \(\mid H _{2}\) \((g,1 \,bar)\) \(\left| H ^{+}( aq ) \| Cu ^{2+}( aq )\right| Cu ( s )\) is \(0.31\, V\). The \(pH\) of the acidic solution is found to be \(3 ,\) whereas the concentration of \(Cu ^{2+}\) is \(10^{- x } \,M\). The value of \(x\) is \(.....\) (Given: \(E _{ Cu ^{2+} / Cu }^{\ominus}=0.34 \,V\) and \(\frac{2.303 RT }{ F }=0.06\, V\) )
- A \(70\)
- B \(7\)
- C \(75\)
- D \(90\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
\(H _{2}( g )+ Cu ^{2+}(\text { aq. }) \rightarrow 2 H ^{+}(\text {aq. })+ Cu ( s )\) \(0.31=0.34-\frac{0.06}{2} \log \frac{\left[ H ^{+}\right]^{2}}{\left[ Cu ^{2+}\right]}\) \({\left[ Cu ^{2+}\right]=10^{-7} M }\) \(x =7\)
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