Chapter 18, Problem 21QP Bookmark Show all steps ON The electrode potential for

Question

Chapter 18, Problem 21QP Bookmark Show all steps ON The electrode potential for the above reaction is given by, 0.0592 Ni lo 4 We have given, solubility product constant of, K for NiPO.-1.7x10-13 This implies, Nij·1.5 × 10-14 272 1.5x Substitute the values of | Ni2. EN/Ni=0.250 V and [Ni]=1 [for solids , concentration is regarded as1] in the equation given below, E =-0.250-0.0592 4 7x10-13(1) When | PO, 1 , E for the reduction of Ni-E" for the given reaction. Thus, substitute the value of P,O1 in the equation (1), we have, E-0.250-0.0592 4 1.7x10-13

Explanation / Answer

Not to worry dear, it is simple typing mistake same value 1.7 × 10^-13 is reappear in equation 1

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