show all workout for electrochemistry problem

Question

Copy and paste link to view the salt bridge graph

https://assets.learnosity.com/organisations/154/e2a8974d-1898-4990-ad79-581c1d41eea7.PNG

Answer the following questions regarding the electrochemical cell shown. A portion of the standard reduction potential table is given below.

L⁺ + e^- → L     E° = 0.20 V

R²⁺ + 2e^- → R   E° = 0.90 V

(a) Write the balanced net-ionic equation for the spontaneous reaction that occurs as the cell operates, and determine the cell voltage. 

(b) In which direction do anions flow in the salt bridge as the cell operates? Justify your answer

(c) If 10.0 mL of a 3.0-molar R(NO₃)₂ solution is added to the half-cell on the right, what will happen to the cell voltage? Explain. 

Answer 1

L+ + e- ==> L Eº = 0.20 V

R²⁺ + 2e- ==> R Eº = 0.90 V

The cathode in this cell will be R and the reduction reduction reaction will take place there. L will be the anode and oxidation will take place there.

(a) R²⁺ + 2e- ==> R

2L ==> 2L⁺ + 2e-

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R²⁺(aq) + 2L ==> 2L⁺⁽aq) + R net ionic

Eºcell = 0.90 V - 0.20 V = 0.70 V

(b) Anions flow into the left compartment containing the L electrode and the 1M L⁺ and NO₃^- because oxidation is taking place here and L is going to L+ so this chamber is gaining + charges. In order to neutralize them, NO3- comes from the salt bridge to maintain electrical neutrality.

(c) The cell voltage will increase

R²⁺(aq) + 2L ==> 2L⁺(aq) + R

From Nernst equation we have...

Ecell = Eº - RT/nF ln Q where Q = [L⁺]² / [R²⁺]

raising the [R²⁺] in the right side will decrease the value of Q making the ln Q negative so Ecell will be greater than Eº