Additional Problems for Chapter 20: What is E_cell of the following voltaic cell

Question

Additional Problems for Chapter 20: What is E_cell of the following voltaic cell? Cu_(s) | Cu^2+ (0.10 M) || Ag_2 CrO_4 (sat'd aq) | Ag_(s) What [Cl^-] should be maintained in the anode half-cell cell is to have E_cell = 0.100 V? K_sp AgCl = 1.8 times 10^-10 Ag, AgCl_(S) | Cl^- (x M) || Cu^2+ (0.25 M) | Cu_(s) For the cell: Ag (s) | Ag^+ (aq, sat'd AgBr) || Ag^+ measured value of E is 0.305 V. What is [Ag^+] in saturated AgBr? What is the experimental value of K_sp for AgBr? Sketch the cell. Determine the potential for the following cell: PbSO_4(s)/SO^2-_4 (0.10 M) || H_3O^+ (1.0 M) | H_2(g)(1.0 at anode is essentially a lead electrode, Pb | Pb^2+(aq). How

Explanation / Answer

Calculation steps _

(i) Determine [ Ag+ ] using the given Ksp value as = 1.1 x 10-12

..........1.1 x 10-12  = [ 2Ag+ ]2 [ CrO42- ]

assuming solubility as x moles / L,

................1.1 x 10-12   = (2x)2 (x)

......................................= 4x3

hence x ie solubility in moles per / L = cube root of (1.1 x 10-12 ) / 4

.............................................................= 0.6503 x 10-4  M

(ii) calculate standard EMF ( Eo ) of the cell , referring to the standard electrode potentials of concerned half cells as ,

..............Cu2+   + 2e -----------> Cu (s ) Eo   = 0.34 V

................Ag+   + e --------------> Ag(s) ; .....Eo   = 0.80 V

...........................Eo = ER   - EL

...................................= ( 0.80 - 0.34 )

..................................= 0.54 V

(iii) Apply Nernst equation to calculate the Ecell of the given voltaic cell

............Ecell   = Eocell   + 2.303 RT / nF{ log [ Cu2+ ] / [ Ag+ ] }

..................or, = Eocell   + { 0.0591 / 1.00 [ log ( 0.1 / 0.6503 x 10-4 ) ] }

........................= 0.54 + { 0.0591 [ -1.0 - ( - 0.1869 + (-4 ) ] }

..........................= 0.54 + { 0.0591 ( -1 + 4.1869) }

..........................= { 0.54 + ( 0.0591 x 3.1869 ) } V

...........................= ( 0.54 + 0.1883 ) V

...........................= 0.7283 V

........................or, = 0.73 V

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Please post the other two questions separately , glad to help.

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