TS EAMCET · Chemistry · Electrochemistry
The emf (in V) of a Daniell cell containing \(0.1 \mathrm{MZnSO}_4\) and \(0.01 \mathrm{M} \mathrm{CuSO}_4\) solutions at their respective electrodes is \(\left(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V} ; E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\right)\)
- A \(1.10\)
- B \(1.16\)
- C \(1.13\)
- D \(1.07\)
Answer & Solution
Correct Answer
(D) \(1.07\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{-}}^{\circ}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ} \\ & =+0.34-(-0.76) \mathrm{V} \\ & =1.1 \mathrm{~V}\end{aligned}\) Further…
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