The standard cell potential ΔE° is 1.100V for the cell Zn(s)/Zn²⁺//Cu²⁺/Cu(s) if [Zn²⁺] = 0.01M and [Cu²⁺] = 1.0M, what is the ΔE or EMF?

To find the cell potential ( E) under non-standard conditions, we use the Nernst equation. Step 1: Write the overall cell reaction and determine the number of electrons transferred (n). The cell notation Zn(s)/Zn^2+//Cu^2+/Cu(s) indicates the following half-reactions: Anode (oxidation): Zn(s) Zn^2+(aq) + 2e^- Cathode (reduction): Cu^2+(aq) + 2e^- Cu(s) Overall reaction: Zn(s) + Cu^2+(aq) Zn^2+(aq) + Cu(s) From the half-reactions, the number of electrons transferred, n, is 2. Step 2: Write the Nernst equation. The Nernst equation at 25^ is: E = E^ - (0.0592)/(n) Q Where E^ is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient. Step 3: Calculate the reaction quotient (Q). For the overall reaction Zn(s) + Cu^2+(aq) Zn^2+(aq) + Cu(s), the reaction quotient is: Q = [Zn^2+][Cu^2+] Given concentrations are [Zn^2+] = 0.01 \, M and [Cu^2+] = 1.0 \, M. Q = (0.01 \, M)/(1.0 \, M) = 0.01 Step 4: Substitute the values into the Nernst equation and solve for E. Given E^ = 1.100 \, V, n=2, and Q=0.01. E = 1.100 \, V - (0.0592)/(2) (0.01) E = 1.100 \, V - (0.0296) (-2) E = 1.100 \, V + 0.0592 \, V E = 1.1592 \, V Rounding to three decimal places, E = 1.159 \, V. Comparing this to the given options: (a) 1.159 \, V (b) 0.923 \, V (c) 1.111 \, V (d) 1.030 \, V (e) 0.991 \, V The calculated value matches option (a). The final answer is (a) 1.159V. Send me the next one 📸