I would choose C bc if u look at it closely u can notice it goes up steadily and then BANG it decreases a lot
Answer:
7.16x10⁻⁸M = [Ag+]
Explanation:
Using the equation:
E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)
<em>Where E</em>⁰<em>= 0.4249V</em>
<em>E(Cell) = -(-0.0019V) -Measured value-</em>
<em>[Cu2+] = 1M</em>
<em />
Replacing:
0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)
-0.423V = - 0.0296 • log (1M/[Ag+]²)
14.29 = log (1M/[Ag+]²)
1.95x10¹⁴ = 1M / [Ag+]²
[Ag+]² = 5.12x10⁻¹⁵M
7.16x10⁻⁸M = [Ag+]
