Question
I have problem solving these problem
Consider the following cell reaction: 2Ag+(aq)+H2(g) rightarrow 2H+(aq)+2Ag(s);E degree Cell = 0.80 V Under standard-state conditions, what is E degree for the following half-reaction? Ag+(aq)+2e- rightarrow Ag(s) 0.40 V - 0.80 V 0.80 V 1.10 V -0.40 V What is the equilibrium constant (K) at 25 degree C for the following cell reaction? Fe(s)+Cd2+(aq) rightarrow Fe2+(aq)+Cd(s), E degree Cell = 0.010V 0.010 0.25 1.0 2.2 1.5 The following cell is initially at standard-state condition. Which of the following statements is true after the cell is allowed to discharge (do work) for a period of time? Zn2+(aq)+2e- Zn(s);E degree = -0.76 V Cu2+(aq)+2e- Cu(s);E degree = 0.34 V Ecell does not change with time. Initially Ecell = - 1.10 V, and it will become more negative with time. Initially Ecell = - 1.10 V, and it will become more positive with time. Initially Ecell = + 1.10 V, and it will becomes more positive with time. Initially Ecell = + 1.10 V, and it will become more negative with time.
Explanation / Answer
1. E0 cell = 0.80/2 V
= 0.40 V as n-factor is halved.
2. E0 = -RTlnK
0.010 = - 8.314 x 298 x lnK
K = 0.99
3. B) is the answer