JEE MAIN - Chemistry (2017 - 9th April Morning Slot - No. 13)
To find the standard potential of M3+/M electrode,the following cell is constituted : Pt/M/M3+(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 M3+ + 3e−$$ \to $$ M at 298 K will be :
(Given $$E_{A{g^ + }\,/\,Ag}^ - $$ at 298 K = 0.80 Volt)
The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3+ + 3e−$$ \to $$ M at 298 K will be :
(Given $$E_{A{g^ + }\,/\,Ag}^ - $$ at 298 K = 0.80 Volt)
0.38 Volt
0.32 Volt
1.28 Volt
0.66 Volt
Explanation
According to Nernst equation, for the cell reaction
M(s) + 3Ag+(aq) $$\to$$ M3+(aq) + 3Ag(s)
$${E_{cell}} = E_{cell}^o - {{0.059} \over n}\log {{[{M^{3 + }}]} \over {{{[A{g^ + }]}^3}}}$$
Substituting the values, we get
$$0.421 = E_{cell}^o - {{0.059} \over 3}\log {{0.001} \over {{{(0.01)}^3}}}$$
$$E_{cell}^o = 0.48V$$
We know that $$E_{cell}^o = E_{A{g^ + }/Ag}^o - E_{{M^{3 + }}/M}^o$$
$$0.48 = 0.8 - E_{{M^{3 + }}/M}^o$$
$$E_{{M^{3 + }}/M}^o = 0.32V$$
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