Given are the following cell which is used to measure pH in solution. Pt, H 2 (1

Question

Given are the following cell which is used to measure pH in solution.

Pt, H2(1 bar) | HCl(aq) || AgCl(s) | Ag

Voltage in this cell at 298.15 is Ecell = 0.517 V

Calculate the pH of the solution.

Hints: You should find [H+] and you need the half-reaction for AgCl:

AgCl(s) + e- --> Ag + Cl-    E0 = 0.22233V

You also need to find out the total reaction in the cell to find Q. The reaction is reduction on AgCl into Ag+ and Cl- and oxidation of H2 into H+. Be aware of coefficient in the chemical equation and amount of electrons.

Explanation / Answer

oxidation: H2---------------------> 2H+ + 2e- , E0 = 0.00V

reduction:

2AgCl(s) + 2e- --------------> 2Ag + 2Cl-    E0 = 0.22233V

Ecell = E0cell -2.303RT/nF* log [H+]^2

Here R= universal gas constant 8.314 J/K mol

T = absolute temperature =25(0C)= 298k

F= faraday = 96500 Coloumb/mol

     n   = no of moles of electrons are transfered =2

2.303RT/F= 0.0591

Ecell = E0cell -(0.0591/n)* log [H+]^2

Ecell = E0cell -(0.0591/2)* log [H+]^2

0.517 =0.22233   - (0.0591)* log [H+]

0.29467 = - (0.0591)* log [H+]

4.99 = -   log [H+]

4.99 = pH

pH = 4.99

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