The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: pt(s)|h2(g,1atm)|h+(aq,1.0m)|au3+(a q,? m)|au(s). what is the concentration of au3+ in the solution if ecell is 1.27 v ? express your answer using two significant figures.

Explanation:

The standard reduction potential is given as 1.27 V.

Concentration of = 1.0 M

Hence, reduction reaction is as follows.

Oxidation reaction is as follows.

Therefore, overall net chemical equation will be as follows.

As, its is known that standard electrode potential () for hydrogen is equal to zero.

And, standard electrode potential for is equal 1.50 V.

Hence, formula to calculate standard cell potential is as follows.

            = -

                               = 1.50 V - 0 V

                                = 1.50 V

Therefore, concentration of is calculated as follows.

Hence, substitute the given values as follows.

                     1.27 = 1.50 - \frac{0.059}{6}log \frac{(1)^{6}}{[Au^{3+}]^{2}}[/tex]

                    = 23

Taking antilog on both the sides, the above equation will be as follows.

                =

                                  =

                  =

                          = M

Hence, we can conclude that the concentration for is M.

idk i’m struggling too

the temperature of the gas, in k, not celsius.