0 votes
35 views
in ElectroChemistry by (330 points)
Calculate the standard electrode potential of the following cell reaction at 25°C: Cu(s) | Cu2+ (0.01 M) || Zn2+ (1.0 M) | Zn(s)

1 Answer

0 votes
by (310 points)
To calculate the standard electrode potential of the given cell reaction, we can use the Nernst equation. First, we need to know the standard reduction potentials for the half-reactions:Cu2+ + 2e-  Cu s   E Cu2+/Cu  = +0.34 VZn2+ + 2e-  Zn s   E Zn2+/Zn  = -0.76 VThe overall cell reaction is:Cu2+ + Zn s   Cu s  + Zn2+Now, we can use the Nernst equation to find the electrode potential  E  at the given concentrations:E = E cell  -  RT/nF  * ln Q where E cell  is the standard cell potential, R is the gas constant  8.314 J/molK , T is the temperature in Kelvin  25C = 298.15 K , n is the number of electrons transferred  2 in this case , F is the Faraday constant  96485 C/mol , and Q is the reaction quotient.First, we need to find E cell :E cell  = E Cu2+/Cu  - E Zn2+/Zn  = 0.34 V -  -0.76 V  = 1.10 VNow, we can calculate Q:Q = [Zn2+]/[Cu2+] =  1.0 M / 0.01 M  = 100Now, we can plug these values into the Nernst equation:E = 1.10 V -  8.314 J/molK * 298.15 K / 2 * 96485 C/mol  * ln 100 E = 1.10 V -  0.0257 V  * ln 100 E = 1.10 V - 0.0257 V * 4.605E  1.10 V - 0.118 VE  0.982 VSo, the standard electrode potential of the given cell reaction at 25C is approximately 0.982 V.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...