Question

3. A student connects a Cd?"(0.20 M)|Cd(s) half-cell to a Cu?"(1 M) Cu(s) electrode. When the red lead is attached to the Cu electrode, the cell potential read by the voltmeter, E is +0.77 V. a. What is the reduction half-reaction at the cathode (red lead)? b. What is the oxidation half-reaction at the anode (black lead)? c. What is the overall cell reaction? d. Write the expression for the thermodynamic reaction quotient, Q, and calculate its value for this cell. e. Use the Nernst equation to find the standard cell potential, E E - Ecall - 0.0257 In Q (at 25°C) f. Knowing that the standard reduction potential, Erd, of the Cu? Cu(s) half-cell is +0.34 V, (or knowing that the standard oxidation potential, E, of the Cu(s)[Cu, half-cell is -0.34 V), what is the potential of the cadmium half-cell? Is this E or Eº? Eºcell - Fºred + E° g. What is the standard reduction potential (SRP, E°) of the Cd? (0.20 M) Cd(s) half-cell? (Hint: This is the value from part f with either the same or opposite sign.)

Answer 1

3 @ - 6 Cd +2 10.2M) | Cliss Cut? (IM) | Curso . (Red lead) Reduction half cell Red lead → Cut lag (IM) | Cuis). Cu trag) + 2e - → Cuis) Oxidiation half cell Black lead Cd Cal te lag, (0.2M). Cd - Cd lagu + 2e- overall celle notations ca Cat2 lag. I cut lag., 1 Cuis). (0.2M 1" (IM) © overall rear 1 Cd, at Cu tag - Cd +2 * Cuis)

n Ecu Eau - 0.0257 en as → 0.77 = Ec-(000287) (2303. dog 10-2) => tru ) 0.77 = Eceu +0.021 Eiew = 0.77-0.021 = 0.749 V - Elen - Ered & Eox. = Ecutallet Ecalcata beregtett = 0.34 F E calca+2 = 0.749 5 Ecclc4+2 = 0.409V] - Cd*210.20M)/ Ca 19) . E poi = E*S*4/4 = -0.4090 (SRP)