A certain half-reaction has a standard reduction potential E⁰ʀᴇᴅ = 0.83 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 0.80V 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 anode of the cell.
(1) Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have? If so, write "yes" and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, write "no".
(2) Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have? If so, write "yes" and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, write "no".

For 1: The minimum standard electrode potential at cathode is -0.03 V

For 2: The maximum standard electrode potential at cathode is 1.63 V

Explanation:

To calculate the of the reaction, we use the equation:

We are given:

= Standard electrode potential of the cell = 0.80 V

To make the standard reduction potential at the cathode minimum, we take the standard reduction potential at anode as (0.83 V)

Putting values in above equation, we get:

Hence, the minimum standard electrode potential at cathode is -0.03 V

To make the standard reduction potential at the cathode maximum, we take the standard reduction potential at anode as (-0.83 V)

Putting values in above equation, we get:

Hence, the maximum standard electrode potential at cathode is 1.63 V

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when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml