1836 United States presidential election in Rhode Island

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1836 United States presidential election in Rhode Island

← 1832 November 3 – December 7, 1836 1840 →
  Van Buren.jpg William Henry Harrison by James Reid Lambdin, 1835 crop.jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Johnson Francis Granger
Electoral vote 4 0
Popular vote 2,964 2,710
Percentage 52.24% 47.76%

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

The 1836 United States presidential election in Rhode Island took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island voted for Democratic candidate Martin Van Buren over Whig candidate William Henry Harrison. Van Buren won Rhode Island by a narrow margin of 4.48%.

This was the first time that Rhode Island ever voted for a Democratic presidential candidate.

Results[edit]

1836 United States presidential election in Rhode Island[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 2,964 52.24% 4 100.00%
Whig William Henry Harrison of Ohio Francis Granger of New York 2,710 47.76% 0 0.00%
Total 5,674 100.00% 4 100.00%

References[edit]

  1. ^ "1836 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.