Question

A certain half-reaction has a standard reduction potential E = -0.45 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.00 V of electrical power. The cell will operate under standard conditions Note for advanced students: assume the engineer requires this half-reaction to happen at the cathode of the cell. 0-0 . " Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? yes, there is a minimum. red - IV If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. no minimum x 6 ? Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? yes, there is a maximum. V If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. TO no maximum By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the anode of this cell. Note: write the half reaction as it would actually occur at the anode.

Answer 1

The question is asked under standard conditions, therefore, E_cell = E^o_cell

Required E_cell = 1.00 V

E^o_cell = E^o_cathode - E^o_anode = 1.00 V (given)

let E^o_anode= x,

-0.45 - x = 1.00

therefore, x = - 1.45

now, the least operating value should be 1.00 V , which means the value shouldn't drop below 1.00 V

so if the the value of cathode drops below - 1.45 V (say -1.46) we will get an output greater than 1.00 V but if it were to rise the value (say -1.44) would be less than 1.00 V [ Rise and fall for negative numbers is opposite]

now for the subpart 1,

There is NO MINIMUM value for E^o_anode.

for subpart 2,

yes, there is a MAXIMUM value. ( if the value exceeds the maximum value, than the cell will operate below 1.00 V)

and that MAXIMUM value = -1.45 V

now lets hop on to subpart 3,

you can use any half cell reaction as long as it is below -1.45 V

so i will take the example of Aluminium (III) ion,

Al³⁺ + 3e^- ------------> Al E^o = -1.676 V

therefore,

E^o_Cell = -0.45 - ( -1.676) = -0.45 + 1.676 = 1.226 V