A voltaic cell consists of a copper electrode in a solution of copper(II) ions w

Question

A voltaic cell consists of a copper electrode in a solution of copper(II) ions with unknown concentration, and a palladium electrode in a 0.1 M solution of palladium(II) ions. The palladium electrode is the cathode and its reduction potential is 0.951 V, the Cu electrode is the anode. (7 points)

(a) Write the half-reaction that occurs at the anode and the cathode, as well as the overall cell reaction.

b) What is the concentration of Cu2+ ions in the anode if the cell potential is 0.584 V? What is the error in the Cu2+ concentration if the cell potential measurement has an error of ±2 mV?

(c) What is the equilibrium constant (Keq) for this reaction?

Explanation / Answer

a)

anode reaction: oxidation takes place

Cu (s) -------------------------> Cu+2 (aq) + 2e-   ,   E0Cu+2/Cu = + 0.340V

cathode reaction : reduction takes palce

Pd+2(aq) + 2e- -----------------------------> Pd(s) , E0Pd+2/Pd = 0.951V

--------------------------------------------------------------------------------

net reaction: Cu(s) +Pd+2(aq) -------------------------> Cu+2 (aq) + Pd(s)

E0cell= E0cathode- E0anode

E0cell= E0Pd+2/Pd - E0Cu+2/Cu

          = 0.951 - (0.340)

          = 0.611V

now

[Pd+2 ] = 0.1 M

[Cu+2 ] = ?

nernest equation

Ecell = E0cell -2.303RT/nF* log [Cu+2]/[Pd+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 [Cu+2]/[Pd+2]

0.584 = 0.611 - (0.059 x1/2) *log [Cu+2 / 0.1]

log [Cu+2 / 0.1] = 0.914

[Cu+2 / 0.1] = 8.2

Cu+2 = 0.82 M

Cu+2 concentration = 0.82 M

error calculation

Cu+2 concentration when potential +2mV = 0.701 M

Cu+2 concentration when potential +2mV = 0.958 M

Error = 0.82 + 0.138

c) equilibrium constant :

E0= 0.059/n *log Kc

0.611 = (0.059 /2 ) x log Kc

logKc = 20.67

Kc = 4.67 x 1020

equilibrium constant = 4.67 x 1020

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