At what $pH$,given half cell $MnO_4^{-} (0.1 \ M) \mid Mn^{2+} (0.001 \ M)$ will have electrode potential of $1.282 \ V$? (Nearest Integer) Given $E_{MnO_4^{-} / Mn^{2+}}^{o} = 1.54 \ V, \frac{2.303 RT}{F} = 0.059 \ V$

  • A
    $3$
  • B
    $2$
  • C
    $1$
  • D
    $6$

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