Question

Hi Expert!

Please respond to all 3 parts:

The half-cells below are connected via a salt-bridge and potentiometer. What is the voltage read by the potentiometer when Aag-2.00 and Ava+-1.50? [[p.2891] Ag+ + 1 e-→ Ag (s) Eo= 0.7993 V va+ + 2 e-→V(s.. Eo =-1.125 V A)1.912 V B)-0.326 V C) 0.313 V D) 937 V E)1.924 V 14-5 E and the Equilibrium Constant A galvanic cell with a large equilibrium constant has a cell potential A) negativ B) constant C) zeo voltage D) positiv忌 E)variable 10. Calculate the equilibrium constant for the reaction between Sn metal and Zn2+ solution. Sn2+ + 2 e-→ Sn (s) E。=-0.141 V Zn2+ + 2 e-→ Zn (s) E° =-0.762 V A) 9.86 × 10 B) .21 x 10-21 C) 3.37 x 1030 D) 3.18 x 10-11 E) 1.84 x 1015

Thank you so much!

Answer 1

Answer – 8 ) We are given the , [Ag⁺] = 2.00 M, [V²⁺] = 1.50

Ag⁺ + 1e^- ----->Ag(s) E^o = 0.7993 V

V²⁺ + 2e^- -----> V(s) E^o = -1.125 V

We know high standard reduction potential means gets reduced .

So, V(s) -----> V²⁺ + 2e^- E^o = 1.125 V

2Ag⁺ + 2e^- ----->Ag(s) E^o = 0.7993 V

2 Ag⁺ (aq) + V(s) -----> V²⁺ + 2Ag(s)   E^ocell = 1.92 V

We know

Ecell = E^ocell – 0.0592/n * log [Ag⁺]² / [V²⁺]

         = 1.92 V – 0.0592/ 2* log 2.00 / (1.50)²

         = 1.925 V

So answer is E) 1.924 V

9) We know in the galvanic cell cell potential are directly proportional to log of equilibrium constant, so as the large equilibrium constants correspond to large positive values

So, A galvanic cell with a large equilibrium constant has a positive cell potential

10) In this one we are given two half reaction

And we know the larger standard reduction potential act ad reduction , so the firs treaction is for reduction and second one for oxidation

Sn²⁺ + 2e^- ------> Sn(s) E^o = -0.141 V

Zn(s) ---> Zn²⁺ + 2e^- E^o = 0.762 V

So overall reaction is

Zn(s) + Sn²⁺ ---> Zn²⁺ + Sn(s) E^o = 0.621 V

We know formula

E^o = (0.0592 V/n) log K

0.621 V = (0.0592V / 2) * log K

0.621 = 0.0296 v * log K

So, log K = 20.97

So K = 9.8*10²⁰

so answer is A) 9.8*10²⁰