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(B) In an alkaline dry cell,the anode reaction involves the oxidation of the metal $(M \rightarrow M^{2+} + 2e^-)$. The standard cell potential $(E^o_

Vedclass · Accepted Answer

Decrease by $0.36 \ V$

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(B) In an alkaline dry cell,the anode reaction involves the oxidation of the metal $(M \rightarrow M^{2+} + 2e^-)$. The standard cell potential $(E^o_{cell})$ is given by $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.Since the cathode remains the same,the change in $E^o_{cell}$ depends on the change in the anode potential $(E^o_{anode})$.For $Zn$,$E^o_{Zn^{2+}/Zn} = -0.76 \ V$.For $Cd$,$E^o_{Cd^{2+}/Cd} = -0.40 \ V$.Replacing $Zn$ with $Cd$ changes the anode potential from $-0.76 \ V$ to $-0.40 \ V$.The new $E^o_{cell}$ will be $E^o_{cathode} - (-0.40 \ V) = E^o_{cathode} + 0.40 \ V$.The original $E^o_{cell}$ was $E^o_{cathode} - (-0.76 \ V) = E^o_{cathode} + 0.76 \ V$.The change in emf is $(E^o_{cathode} + 0.40) - (E^o_{cathode} + 0.76) = -0.36 \ V$.Therefore,the standard emf of the cell decreases by $0.36 \ V$.

Suppose that an alkaline dry cell was manufactured using cadmium metal rather than $Zn$. What effect would this have on the standard emf of the cell? The standard reduction potential $E^o_{Cd^{2+}/Cd} = -0.40 \ V$ while $E^o_{Zn^{2+}/Zn} = -0.76 \ V$.

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