Electro Chemistry

1.10 V

Step 1: In the Daniel cell, Zn is the anode (oxidation) and Cu is the cathode (reduction). Step 2: Standard oxidation potential of Zn = −(−0.76) = +0.76 V (reverse the sign of reduction potential). Step 3: Standard reduction potential of Cu²⁺/Cu = +0.34 V. Step 4: E°cell = E°ox(anode) + E°red(cathode) = 0.76 + 0.34 = 1.10 V. Option B (0.42) is obtained by subtracting, which is wrong here. Option C gives a negative value which would indicate a non-spontaneous reaction. Option D only uses Zn's value and ignores Cu.

Number of moles of electrons transferred in the balanced cell reaction

Step 1: The Nernst equation is derived from ΔG = −nFEcell and ΔG = ΔG° + RT ln Q. Step 2: Here 'n' is the number of moles of electrons exchanged between the oxidising and reducing agents in the overall balanced cell reaction. Step 3: For the Daniel cell (Zn + Cu²⁺ → Zn²⁺ + Cu), n = 2 because 2 electrons are transferred. Step 4: This 'n' appears because Faraday's law uses nF as total charge (n moles of electrons × F coulombs/mol). Other options describe different quantities entirely and would change the physical meaning of the equation.

2.68 g

Step 1: Calculate charge Q = I × t = 2 A × (20 × 60) s = 2 × 1200 = 2400 C. Step 2: Find electrochemical equivalent Z = Molar mass / (n × F) = 108 / (1 × 96500) g/C. (n=1 because Ag⁺ + e⁻ → Ag requires 1 electron). Step 3: Use m = Z × I × t = Z × Q = (108/96500) × 2400. Step 4: m = (108 × 2400) / 96500 = 259200 / 96500 ≈ 2.68 g. Option B would be double (wrong time/current), Option C is half the correct answer, Option D is the molar mass itself (deposited only when 1 mole = 96500 C is passed).

Na⁺(l) + e⁻ → Na(l)

Step 1: In an electrolytic cell, the cathode is connected to the negative terminal of the power supply. Step 2: The cathode attracts cations (positive ions) from the melt. Step 3: Na⁺ ions are attracted to the cathode and gain electrons — this is reduction. Step 4: Na⁺(l) + e⁻ → Na(l), E° = −2.71 V. Option B is the anode reaction (oxidation of Cl⁻). Option C is also an oxidation reaction, which occurs at the anode. Option D is reduction of Cl₂, which does not happen under these conditions. Cathode = reduction = gain of electrons.