JEE Main 2017 · 9 Apr · Shift 1 · Chemistry Q36

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ChemistrySingle correct+4 / -1Question 36 of 90

To find the standard potential of $\ce{M^3+/M}$ electrode, the following cell is constituted : $\ce{Pt/M/M^3+}$ (0.001 mol L $^{-1}$ )/ $\ce{Ag+}$ (0.01 mol L $^{-1}$ )/Ag

The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction $\ce{M^3+ + 3e- -> M}$ at 298 K will be :

(Given $E^\circ_{\ce{Ag+/Ag}}$ at 298 K = 0.80 Volt)

0.38 Volt

0.32 Volt

1.28 Volt

0.66 Volt

Answer and worked solution

Official answer: B

The cell reaction is $\ce{M + 3Ag+ -> M^3+ + 3Ag}$ , so
$E=E^\circ_{cell}-\frac{0.059}3\log\frac{10^{-3}}{(10^{-2})^3}=E^\circ_{cell}-0.059.$
$E^\circ_{cell}=0.48$ V $=0.80-E^\circ_{\ce{M^3+/M}}$ , giving $E^\circ=0.32$ V.

Topics:Nernst equation Galvanic cells and EMF

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JEE Main 2017 · 9 Apr · Shift 1, Chemistry question 36