ExamBro
ExamBro
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 \(\left(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V} ; E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\right)\)

  1. A \(1.10\)
  2. B \(1.16\)
  3. C \(1.13\)
  4. D \(1.07\)
Verified Solution

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…