1836 United States presidential election in Rhode Island

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1836 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1832 November 3 – December 7, 1836 1840  
  Martin Van Buren circa 1837 crop.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 vote40
Popular vote2,9642,710
Percentage52.24%47.76%

Rhode Island Presidential Election Results 1836.svg
County Results

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.

Contents

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, and Van Buren's performance would not be bettered by a Democrat in Rhode Island until Franklin D. Roosevelt in 1932. [1]

Results

1836 United States presidential election in Rhode Island [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 2,96452.24%4100.00%
Whig William Henry Harrison of Ohio Francis Granger of New York 2,71047.76%00.00%
Total5,674100.00%4100.00%

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References

  1. "Presidential General Election Results Comparison – Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved October 25, 2019.
  2. "1836 Presidential General Election Results - Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved December 23, 2013.