1836 United States presidential election in Ohio

A presidential election was held in Ohio on November 4, 1836 as part of the 1836 United States presidential election.^[1] Voters chose twenty-one representatives, or electors to the Electoral College, who voted for President and Vice President.

Quick facts Nominee, Party ...

1836 United States presidential election in Ohio

← 1832 November 4, 1836 1840 â†’

PA â†’

Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate Francis Granger Richard Mentor Johnson Electoral vote 21 0 Popular vote 104,958 96,238 Percentage 51.87% 47.56%

County Results Harrison   50–60%   60–70%   70–80% Van Buren   50–60%   60–70%   70–80% No Votes   No data

President before election Andrew Jackson Democratic Elected President Martin Van Buren Democratic

Close

Ohio voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Ohio by a narrow margin of 4.31%. Ohio was the home state of William Henry Harrison.