Aconcentration cell is built based on the following half reactions by using two pieces of zinc as electrodes, two zn2+ solutions, 0.129 m and 0.427 m, and all other materials needed for a galvanic cell. what will the potential of this cell be when the cathode concentration of zn2+ has changed by 0.047 m at 291 k?
zn2+ + 2 e- ? zn eo = -0.761 v

Explanation:

The given reaction at cathode will be as follows.

At cathode: ,     = -0.761 V

At anode: ,       = 0.761

Therefore, net reaction equation will be as follows.

Initial:     0.129         -            -       0.427

Change:  -0.047      -            -     -0.047

Equilibrium: (0.129 - 0.047)      (0.427 - 0.047)

                = 0.082                       = 0.38

As for the given reaction is zero.

Hence, equation for calculating new cell potential will be as follows.

                   E_{cell} =

                              =

                              = 0.019

Thus, we can conclude that the cell potential of the given cell is 0.019.

its going to be mutualism cuz they both benefit from the relationship.

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the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l