The change in electrode potential of Cr^{3+}/ | vedclass

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(B) The half-cell reaction is: $2Cr^{3+}{_{	ext{(aq)}}} + 7H_2O_{	ext{(l)}} ightarrow Cr_2O_7^{2-}{_{	ext{(aq)}}} + 14H^{+}{_{	ext{(aq)}}} + 6e^

Vedclass · Accepted Answer

$0.28$

Vedclass · Answer

(B) The half-cell reaction is: $2Cr^{3+}{_{	ext{(aq)}}} + 7H_2O_{	ext{(l)}} ightarrow Cr_2O_7^{2-}{_{	ext{(aq)}}} + 14H^{+}{_{	ext{(aq)}}} + 6e^{-}$
Using the Nernst equation: $E = E^o - \frac{0.06}{6} \log \frac{[Cr_2O_7^{2-}][H^{+}]^{14}}{[Cr^{3+}]^2}$.
$E = E^o - 0.01 \log \frac{[Cr_2O_7^{2-}]}{[Cr^{3+}]^2} - 0.01 	imes 14 \log [H^{+}]$.
Since $pH = -\log[H^{+}]$,we have $E = E^o - 0.01 \log \frac{[Cr_2O_7^{2-}]}{[Cr^{3+}]^2} + 0.14 \ pH$.
The change in potential $\Delta E = E_2 - E_1 = 0.14(pH_2 - pH_1)$.
Given $pH_1 = 1$ and $pH_2 = 3$,$\Delta E = 0.14(3 - 1) = 0.14 	imes 2 = 0.28 \ V$.

$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$	$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$	$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

The change in electrode potential of $Cr^{3+}/Cr_2O_7^{2-}$ electrode at $25 \ ^oC$ due to change in $pH$ of its electrolytic solution from $1$ to $3$ is ........... $V$ (Assume: $[Cr_2O_7^{2-}]$ and $[Cr^{3+}]$ remain constant). Use: $\frac{2.303RT}{F} = 0.06$.

Similar Questions

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