Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?

 
 Zn =⇒ Zn+2(0.10) + 2e- (anode)
 Zn+2(?M) + 2e- === Zn(s) (cathode)
 
 Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn 
 E = E^o -0.0592 log Q; in this case E^o is zero. 
 E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2 
 23 mV x 1 volt/1000mv = 0.023 Volts 
 0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
 0.023 = -0.0296 { log 0.10 – log [Zn+2] }
 0.023 = -0.0296{ -1 - log[Zn+2] }
 0.023 = +0.0296 + 0.0296log[Zn+2]
 -0.0066 = 0.0296log[Zn+2]
 -0.22= log[Zn+2]
 [Zn+2] = 10^-0.22 = 0.603 Molar

Yes she is correct. the answer is completely correct

500bc indiais the answer.