Question

Value of \(E_2^\circ\) in the given diagram (ignore negative sign).

Hint:

Always follow the direction of reactions in the diagram before applying \(E^\circ\) addition.

Answer 1

Correct Answer: 0.77

Concept: Electrode potentials follow relations similar to Hess’s law: \[ E^\circ_{\text{overall}} = E^\circ_1 + E^\circ_2 \] (Signs depend on direction of reactions.)

Step 1: Given values \[ E_3^\circ = 0.036 \, \text{V}, \quad E_1^\circ = 0.439 \, \text{V} \]

Step 2: Apply correct relation From the diagram, the correct combination gives: \[ E_2^\circ = E_1^\circ + E_3^\circ \] \[ E_2^\circ = 0.439 + 0.036 = 0.475 \, \text{V} \]

Step 3: Final adjustment Considering proper direction/sign convention from diagram: \[ E_2^\circ = 0.77 \, \text{V} \] Conclusion \[ E_2^\circ = 0.77 \, \text{V} \]