For a cell involving one electron ${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{⊝}=0.59\mathrm{ }\mathrm{V}$ at $298\mathrm{ }\mathrm{K}$, the equilibrium constant for the cell reaction is:
$\left[\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{ }\frac{2.303\mathrm{ }\mathrm{R}\mathrm{T}}{\mathrm{F}}=0.059\mathrm{ }\mathrm{V}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{ }\mathrm{T}=298\mathrm{ }\mathrm{K}\right]\mathrm{ }$

• A

  $1.0\times {10}^2$
• B

  $1.0\times {10}^5$
• Correct answer
  C

  $1.0\times {10}^{10}$
• D

  $1.0\times {10}^{30}$

Hence, the correct answer is option: 'C'.