An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)₂(s) + 2e^- Fe(s) + 2OH^-(aq) E^ = -0.88 V and AgBr(s) + e^- Ag(s) + Br^-(aq) E^ = +0.07 V Which of the following option is correct ?

Step-by-step solution

For a spontaneous reaction, the cell potential E^ _ cell must be positive. The half-cell with the higher standard reduction potential acts as the cathode (reduction), and the one with the lower standard reduction potential acts as the anode (oxidation). Given standard reduction potentials: E^ _ AgBr/Ag = +0.07 V E^ _ Fe(OH)₂/Fe = -0.88 V Since +0.07 V > -0.88 V, AgBr undergoes reduction at the cathode and Fe undergoes oxidation at the anode. Cathode reaction: 2AgBr(s) + 2e⁻ 2Ag(s) + 2Br⁻(aq) Anode reaction: Fe(s) +