We need to use the Cr-Au galvanic cell to power a 75.0-watt light bulb. Please determine how long (in min) the galvanic cell will power the light bulb before running out power. The galvanic cell uses the following solutions: 0.810 L 1.50 M Cr2(SO4)3 solution and 1.20 L 0.860 M Au(NO3)3 solution. power(W) = electric potential(V) x current(A); charge(C)= current(A) x time(s); 1 mol of electrons = 96485 C

148.6 mins

Explanation:

Aim: using Cr-Au galvanic cell to power a 75 watt light bulb

Calculate how long the Galvanic cell will power the light bulb

power ( W ) = electric potential * current   ( 1 )

charge = current * time ( 2 )

first step : determine the electric potential

E⁰cell = E⁰cathode  -  E⁰anode ( values gotten from balancing  reaction )

          = ( 1.5 - ( -0.74 )) =  2.24 V

Given that amount of AU^+3 is limited the

Ecell ( cell voltage ) ≈ E⁰cell  =  2.24 V

back to equation 1

Power = 2.24 * current

∴ current = 75 / 2.24 = 33.48 A

back to equation 2

Charge = current * time

Time = Charge / current  = ( 3 * 1.032 * 96485 ) / (33.48 * 60 )

                                         = 148.6 mins

Note : 1 mol AU^+3  will give  3 e^-

          1.032 mol AU^+3 = 3 * 1.032 mol e^-

hence charge = 3 * 1.032 * 96485

answer: 500 wph.

explanation: