The electrode potential of the following half cell at $298 \ K$ is given by the cell reaction:
$X | X^{2+}(0.001 \ M) || Y^{2+}(0.01 \ M) | Y$
The cell potential is $....... \times 10^{-2} \ V$ (Nearest integer).
Given: $E^0_{X^{2+} | X} = -2.36 \ V$,$E^0_{Y^{2+} | Y} = +0.36 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$.

  • A
    $274$
  • B
    $273$
  • C
    $272$
  • D
    $275$

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To find the standard potential of $M^{3+}/M$ electrode,the following cell is constituted:
$Pt | M | M^{3+} (0.001 \ mol \ L^{-1}) || Ag^{+} (0.01 \ mol \ L^{-1}) | Ag$
The $emf$ of the cell is found to be $0.421 \ V$ at $298 \ K$. The standard potential of the half-reaction $M^{3+} + 3e^{-} \to M$ at $298 \ K$ will be .............. $V$.
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The potential for the given half cell at $298 \ K$ is $(-) \ldots \ldots \ldots \times 10^{-2} \ V.$
$2 H^{+}_{(aq)} + 2 e^- \rightarrow H_{2(g)}$
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