What is the e.m.f. of the cell in V?Cr | Cr^{ | vedclass

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(C) The cell reaction is: $2Cr + 3Co^{2+} \rightarrow 2Cr^{3+} + 3Co$.Since the concentrations are $1.0 \ M$,the reaction quotient $Q = 1$,and $\ln(Q)

Vedclass · Accepted Answer

$+0.46$

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(C) The cell reaction is: $2Cr + 3Co^{2+} \rightarrow 2Cr^{3+} + 3Co$.Since the concentrations are $1.0 \ M$,the reaction quotient $Q = 1$,and $\ln(Q) = 0$.Using the formula: $E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log(Q)$.Since $Q = 1$,$E_{cell} = E_{cell}^{o}$.$E_{cell}^{o} = (E_{RP}^{o})_{cathode} - (E_{RP}^{o})_{anode}$.Here,the cathode is $Co^{2+} | Co$ and the anode is $Cr | Cr^{3+}$.$E_{cell}^{o} = (-0.28 \ V) - (-0.74 \ V) = +0.46 \ V$.

What is the $e.m.f.$ of the cell in $V$?
$Cr | Cr^{3+} (1.0 \ M) || Co^{2+} (1.0 \ M) | Co$
Given: $E^{o}$ for $Cr^{3+} | Cr = -0.74 \ V$ and $Co^{2+} | Co = -0.28 \ V$

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What is the $e.m.f.$ of the cell in $V$?$Cr | Cr^{3+} (1.0 \ M) || Co^{2+} (1.0 \ M) | Co$Given: $E^{o}$ for $Cr^{3+} | Cr = -0.74 \ V$ and $Co^{2+} | Co = -0.28 \ V$