Half cell reactions Oxidation half cell(Anode) Ca(s) -------> Ca 2+ (aq) + 2e E°

Question

Half cell reactions

Oxidation half cell(Anode)

Ca(s) -------> Ca2+(aq) + 2e E°red= -2.87V

Reduction half cell(cathode)

Mn2+(aq) + 2e --------> Mn(s) E°red = -1.18V

short hand notation

Ca(s)|Ca2+(aq)||Mn2+(aq)|Mn(s)

Standard cell potential(E°)

E°cell = E°red,Cathode - E°red,anode

     = -1.18 V - (-2.87V)

= -1.18 + 2.87V

= 1.69V

Standard free Energy Change(G°)

G° = -nFE°

= - (2 × 964850C/mol × 1.69V)

= 326.12kJ/mol

Equillibrium Constant(K)

G° = - RTlnK

logK = -G°/(2.303 RT)

= - (-326.12(kJ/mol)/2.303×0.008314kJ/mol K × 298.15K)

= 57.13

K = 1.3 ×10^57

Explanation / Answer

8 Write the half-cel occurs at 26 C l equations and the short hand notation representing the following a voltaic cell reaction Cals) + Mn"(aq)(1 M) Ca"(aq(1 M) + Mn(s). Half-cell equations [2 points] Short-hand Notation [2 points]: Also, calculate the standard cell potential, the free energy in kilojoules and the the information given below 18 AG=-mFE', where Fs 965001/v.mol e. AG--RTInKeq, where R-8.3141/K E Cell: 12 points): AG[2 points): Keg [3 points):

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