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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

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 \(0.421\, volt\) at \(298\, K\). The standard potential of half reaction \(M^{3+} + 3e \to M\) at \(298\, K\) will be .............. \(\mathrm{volt}\) (Given \(E_{A{g^ + }/Ag}^o \) at \(298\, K\, = 0.80\, Volt\) )

  1. A \(0.38\)
  2. B \(0.32\)
  3. C \(1.28\)
  4. D \(0.66\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.32\)

Step-by-step Solution

Detailed explanation

Cell reaction : \(M\, + \,3A{g^ + }(aq)\, \to \,{M^{3 + }}\, + \,3Ag(s)\) \(E\, = \,{E^o}\, - \,\frac{{0.0591}}{n}\log \,\) \(\frac{{[\operatorname{Re} duction\,\,state]}}{{[Oxidised\,\,\,state]}}\)…
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