




Reaction mechanism: Lithium borohydride (LiBH$_4$) is a selective reducing agent that reduces esters (CO$_2$Et) to alcohols while leaving carboxylic acids (CO$_2$H) unchanged.
Step-by-step process:
Option (A) The ester group (CO$_2$Et) is reduced by LiBH$_4$ in ethanol (EtOH) to form a primary alcohol (-CH$_2$CH$_2$OH).
Option (B) The carboxylic acid group (CO$_2$H) remains unchanged during the reaction.
Option (C) Protonation with H$_3$O$^+$ ensures the stability of the final product.
The final product contains an unchanged carboxylic acid group and a newly formed primary alcohol group, corresponding to option (4).
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).

Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to
