The magnitudes (but not the signs) of the standard reduction potentials of two metals \(X\) and \(Y\) are $$ \begin{array}{ll} \mathrm{Y}^{2+}+2 e^{-} \longrightarrow \mathrm{Y} & \left|E^{\circ}\right|=0.34 \mathrm{~V} \\ \mathrm{X}^{2+}+2 e^{-} \longrightarrow \mathrm{X} & \left|E^{\circ}\right|=0.25 \mathrm{~V} \end{array} $$ where the \(\|\) notation denotes that only the magnitude (but not the sign) of the \(E^{\circ}\) value is shown. When the half-cells of \(X\) and \(Y\) are connected, electrons flow from \(X\) to \(Y\). When \(X\) is connected to a SHE, electrons flow from \(X\) to SHE. (a) Are the \(E^{\circ}\) values of the half- reactions positive or negative? (b) What is the standard emf of a cell made up of \(X\) and \(Y ?\)

Step by step solution

01

Determining the Signs of Standard Reduction Potentials

Since electrons flow from X to Y when their half-cells are connected, Y must be accepting them and is thus the cathode. Reduction takes place at the cathode, so \(Y^{2+}\) is being reduced and its reduction potential must be positive i.e. \(E^{\circ}_Y=+0.34\) V. Similarly, since electrons flow from X to SHE, SHE is accepting them and is the cathode. But we know that the reduction potential of SHE is 0 V. Thus, the reduction potential of X must be negative, or \(E^{\circ}_X=-0.25\) V.

02

Calculating Standard Emf of the Cell

We can now calculate the standard emf of a cell made up of X and Y using the formula \(E^{\circ}_{cell}=E^{\circ}_cathode-E^{\circ}_anode\). Here, the cathode is Y and the anode is X, so \(E^{\circ}_{cell}=E^{\circ}_Y-E^{\circ}_X=+0.34-(-0.25)=0.59\) V.

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