Calculate the standard emf for each of the following reactions.
h2(g)+f2(g)→2h+(aq)+2f−(aq).
cu2+(aq)+ca(s)→cu(s)+ca2+(aq).

In tables, all reaction are reduction reactions For H : 2 H+ + 2 e- -> H2 E = 0 V (definition) In first reaction H oxidized, and E = 0 V Then find E for reaction F2 + 2e -> 2 F- calculate sum together In next reaction Copper is reduced and Ca oxidized Find E for reactions Cu2+ + 2e- -> Cu Ca2+ + 2e- -> Ca Because lower reaction happens from right to left, 

Ithink is a because if say increased in electronegativity