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GeneralMedium JEE chemistry MCQ

For the following electrochemical cell at $298 \mathrm{~K}$, $\mathrm{Pt}(s) \mid \mathrm{H}_{2}(g, 1$ bar $)\left|\mathrm{H}^{+}(aq, 1 \mathrm{M}) \| \mathrm{M}^{4+}(aq), \mathrm{M}^{2+}(aq)\right| \operatorname{Pt}(s)$ $E_{\text {cell }}=0.092 \mathrm{~V}$ when $\frac{\left[\mathrm{M}^{2+}(aq)\right]}{\left[M^{4+}(aq)\right]}=10^{x}$. Given : $E_{\mathrm{M}^{4} / \mathrm{M}^{2+}}^{0}=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}$ The value of $x$ is
  1. A. -2
  2. B. -1
  3. C. 1
  4. D. 2

Solution

The correct option is **D**.

CHEMISTRY

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Question
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For the following electrochemical cell at 298 K298 \mathrm{~K}, Pt(s)H2(g,1\mathrm{Pt}(s) \mid \mathrm{H}_{2}(g, 1 bar )H+(aq,1M)M4+(aq),M2+(aq)Pt(s))\left|\mathrm{H}^{+}(aq, 1 \mathrm{M}) \| \mathrm{M}^{4+}(aq), \mathrm{M}^{2+}(aq)\right| \operatorname{Pt}(s) Ecell =0.092 VE_{\text {cell }}=0.092 \mathrm{~V} when [M2+(aq)][M4+(aq)]=10x\frac{\left[\mathrm{M}^{2+}(aq)\right]}{\left[M^{4+}(aq)\right]}=10^{x}. Given : EM4/M2+0=0.151 V;2.303RTF=0.059 VE_{\mathrm{M}^{4} / \mathrm{M}^{2+}}^{0}=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V} The value of xx is
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General — Medium JEE Chemistry MCQ | MyGoalPrep