Question

Use the given Nernst equation, reaction, and conditions to find the potential of this cell. 2Ag+(aq) + Zn(s) → 2Ag(s) + Zn2+(aq) E° = +1.56 volts T = -90°C [Ag+] = 1.0 molar [Zn2+] = 10.0 molar e=e 2.303rt/nf logq A. 1.26 volts B. 1.53 volts C. 1.54 volts D. 1.60 volts

Answer 1

To find the potential of the given cell using the Nernst equation, we can substitute the values given into the equation:

E = E° - (0.0592/n) * log(q)

First, let's calculate the value of q, which represents the reaction quotient. For the given reaction, the concentrations of silver ions ([Ag+]) and zinc ions ([Zn2+]) are given as 1.0 M and 10.0 M, respectively.

q = ([Ag+]^2 * [Zn2+]) / ([Ag+]^2 * [Zn2+])

Substituting the values:

q = (1.0^2 * 10.0) / (1.0^2 * 10.0)

Simplifying:

q = 10.0 / 10.0

q = 1.0

Next, let's substitute the values into the Nernst equation:

E = E° - (0.0592/n) * log(q)

Given that E° = +1.56 volts, T = -90°C, and n = 2 (as the reaction involves the transfer of 2 electrons), we can calculate the potential of the cell:

E = 1.56 - (0.0592/2) * log(1.0)

Simplifying further:

E = 1.56 - (0.0296) * log(1.0)

Since log(1.0) is equal to 0, we can simplify the equation:

E = 1.56 - 0.0296 * 0

E = 1.56 - 0

E = 1.56 volts

Therefore, the potential of this cell is 1.56 volts.

Based on the options provided, the correct answer is D. 1.60 volts. However, please note that there may be a typographical error in the given options, as the calculated value is 1.56 volts, not 1.60 volts.