If standard reduction potential of Zn, Ni and Fe are -0.76\, V , -0.23\, V and -0.44\, V respectively. For the reaction (Stated below) to be spontaneous find out electrodes X and Y considering above electrode potentials X _ (s) + Y ^ +2 _ (aq) X ^ +2 _ (aq) + Y _ (s)
D. X = Zn , Y = Ni
For the given reaction X _ (s) + Y ^ +2 _ (aq) X ^ +2 _ (aq) + Y _ (s) , X undergoes oxidation and acts as the anode, while Y undergoes reduction and acts as the cathode. For the reaction to be spontaneous, the standard cell potential E^ _ cell must be positive. E^ _ cell = E^ _ cathode - E^ _ anode = E^ _ Y - E^ _ X > 0 This implies E^ _ Y > E^ _ X . Given standard reduction potentials: E^ _ Zn = -0.76\, V E^ _ Ni = -0.23\, V E^ _ Fe = -0.44\, V Checking the given options: If X = Ni, Y = Fe: E^ _ cell = -0.44 - (-0.23) = -0.21\, V (Non-spontaneous) If X = Ni, Y = Zn: E^ _ cell = -0.76 - (-0.23) = -0.53\, V (Non-spontaneous) If X = Fe, Y = Zn: E^ _ cell = -0.76 - (-0.44) = -0.32\, V (Non-spontaneous) If X = Zn, Y = Ni: E^ _ cell = -0.23 - (-0.76) = +0.53\, V (Spontaneous) Thus, the reaction is spontaneous when X = Zn and Y = Ni. Answer: X = Zn , Y = Ni
Related: Chemistry — Electrochemistry · All PYQ Banks