Chapter 17, Problem 101QAP

Use Table 17.1 to answer the following questions. Use LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required).

(a) For the half reaction $\frac{1}{2}{\text{Br}}_2\left(l\right)+e^{-}\to {\text{Br}}^{-}\left(aq\right)$   $E_{\text{red}}^{°}$1.077 v.

(b) For the reaction $2{\text{Br}}^{-}\left(aq\right)+{\text{Co}}^{2+}\left(aq\right)\to {\text{Br}}_2\left(l\right)+\text{Co}\left(s\right)$   $E°$0.

(c) If the half reaction ${\text{Co}}^{2+}\left(aq\right)+2e^{-}\to \text{Co}\left(s\right)$ is designated as the new standard where $E_{\text{red}}^{°}$ is 0.00, then $E_{\text{red}}^{°}$ for $2{\text{H}}^{+}\left(aq\right)+2e^{-}\to {\text{H}}_2\left(g\right)$ is 0.

(d) For the reaction $2{\text{Cr}}^{3+}\left(aq\right)+3\text{Co}\left(s\right)\to 2\text{Cr}\left(s\right)+3{\text{Co}}^{2+}\left(aq\right)$ the number of electrons exchanged is6.

(e) For the reaction described in (d), the number of coulombs that passes through the cell is $9.648\times {10}^4$ .