Calculate the emf of the cell: Cr | Cr^{3+}(0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The standard cell potential $E^0_{cell}$ is calculated as $E^0_{cathode} - E^0_{anode}$.$E^0_{cell} = E^0_{Fe^{2+}/Fe} - E^0_{Cr^{3+}/Cr} = -0.45

Vedclass · Accepted Answer

$0.2606 \ V$

Vedclass · Answer

(A) The standard cell potential $E^0_{cell}$ is calculated as $E^0_{cathode} - E^0_{anode}$.$E^0_{cell} = E^0_{Fe^{2+}/Fe} - E^0_{Cr^{3+}/Cr} = -0.45 \ V - (-0.75 \ V) = 0.30 \ V$.The Nernst equation at $298 \ K$ is $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$.Here,$n = 6$ (number of electrons transferred).$E_{cell} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.01)^3} = 0.30 - 0.00985 \log \frac{10^{-2}}{10^{-6}} = 0.30 - 0.00985 \log(10^4)$.$E_{cell} = 0.30 - 0.00985 	imes 4 = 0.30 - 0.0394 = 0.2606 \ V$.

Calculate the $emf$ of the cell: $Cr | Cr^{3+}(0.1 \ M) || Fe^{2+}(0.01 \ M) | Fe$. Given: $E^0_{Cr^{3+}/Cr} = -0.75 \ V$; $E^0_{Fe^{2+}/Fe} = -0.45 \ V$. Cell reaction: $2 \ Cr_{(s)} + 3 \ Fe^{2+}_{(aq)} \rightarrow 2 \ Cr^{3+}_{(aq)} + 3 \ Fe_{(s)}$.

Similar Questions

What will be the electromotive force of the following cell $Fe|Fe^{+2}\,(0.2\, M)\,||\, Au^{+3}\,(0.02\, M)\,|Au$ (in $,V$)?
Given: $E^o_{Fe^{+2}/Fe} = -0.44\,V$ and $E^o_{Au^{+3}/Au} = +1.50\,V$.

Which one of the following has a potential more than zero?

If the $E^{\circ}_{cell}$ for a given reaction has a negative value,which of the following gives the correct relationships for the values of $\Delta G^{\circ}$ and $K_{eq}$ ?

Give the Nernst equation for the following reaction: $M_{(aq)}^{n+} + ne^{-} \to M_{(s)}$

If a solution of $Cu^{+2}/Cu$ at $298 \, K$ is diluted $100$ times,how will the electrode potential change?

Vedclass Test Series

Exam Paper Generator

Online Exam Module