What is the pH of the solution in the cathode compartment of the following cell

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Zn(s) ---> Zn2+ & 2 e- Eo = + 0.763 volts 2 H+ 2 e- ---> H2(1atm) Eo = 0 volts adding the 2 equations we get Zn(s+ 2 H+1 ==== H2 + Zn+2 E0 cell = E0 oxiddation+E reduction= .763+0 = .763 Now Applying Nerst Equation E cell = E0 cell - .06/n log K here K is the equilibrium constant E cell= E0 cell-.06/2 log ([Zn+2][H2]/[H+1]^2)..here H2 = 1 atm STP zn+2 = 1 M so E cell = .763 - .03log(zn+2/H+1^2) 0.763-0.51=.03log(zn+2/H+1^2) 8.433= log(zn+2/H+1^2) H+1 = 6.07 * 10-5 PH= -log[H+] PH=4.21

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